The composition of functions fulfills the associativity property, i.e. if $f:A\mapsto B,$ $g:B\mapsto C,$ and $h:C\mapsto D$ are functions then $(h\circ g)\circ f=h\circ (g\circ f).$
new:branchproof: * Let $f:A\mapsto B,$ $g:B\mapsto C,$ and $h:C\mapsto D$ be functions. * The lemma defining compositions shows that both compositions $(h\circ g)\circ f$ and $h\circ (g\circ f)$ are functions. * Moreover, these functions are identical, since for every $a\in A$ we have $$(h\circ g)\circ f(a)=h(g\circ f(a))=h(g(f(a)))=h\circ g(f(a))=h\circ(g\circ f)(a).$$
Proofs: 1
